# How do you solve using the completing the square method x^2-6x+5=0?

Jul 15, 2016

$\textcolor{b l u e}{x = 5}$ or $\textcolor{b l u e}{x = 1}$

#### Explanation:

Remember in squaring a general binomial
$\textcolor{w h i t e}{\text{XXX}} {\left(a + b\right)}^{2} = {a}^{2} + 2 a b + {b}^{2}$

Given: ${x}^{2} - 6 x + 5 = 0$

If ${x}^{2}$ and $- 6 x$ are the first two terms of the expansion of a squared binomial:
$\textcolor{w h i t e}{\text{XXX}} a = 1$, and
$\textcolor{w h i t e}{\text{XXX}} b = - 3$

So the third term must be ${b}^{2} = {\left(- 3\right)}^{2} = 9$

To complete the square we add a $9$ to the first two terms (but to keep everything balanced we will need to subtract it again:
$\textcolor{w h i t e}{\text{XXX}} {x}^{2} - 6 x \textcolor{red}{+ 9} + 5 \textcolor{red}{- 9} = 0$

Now we can write the first 3 terms as a squared binomial:
$\textcolor{w h i t e}{\text{XXX}} {\left(x - 3\right)}^{2} + 5 - 9 = 0$

Simplifying $\left(5 - 9\right) = - 4$ and adding $4$ to both sides gives
$\textcolor{w h i t e}{\text{XXX}} {\left(x - 3\right)}^{2} = 4$

Taking the square root of both sides
$\textcolor{w h i t e}{\text{XXX}} x - 3 = \pm 2$

Then add $3$ to both sides
$\textcolor{w h i t e}{\text{XXX}} x = 3 \pm 2$

$\Rightarrow x = 5 \mathmr{and} x = 1$

Jul 15, 2016

$x = 5 , \text{ or } x = 1$

#### Explanation:

It seems strange to want to use completing the square, because this quadratic trinomial factorises to give $\left(x - 5\right) \left(x - 1\right) = 0$

But let's proceed... Completing the square is based on the fact that
the square of a binomial gives a standard answer

${\left(x - 8\right)}^{2} = {x}^{2} - \textcolor{b l u e}{16} x + \textcolor{m a \ge n t a}{64}$

There is ALWAYS a relationship between the $\textcolor{b l u e}{16} \mathmr{and} \textcolor{m a \ge n t a}{64}$

This is: $\textcolor{b l u e}{16 \div 2} \text{ and then squared gives "color(magenta)64 " } {8}^{2} = 64$

In ${x}^{2} - 6 x + 5 = 0 \text{ the relationship is not correct}$

This means that 5 is not the correct constant, move it to the RHS

${x}^{2} - \textcolor{b l u e}{6} x \textcolor{m a \ge n t a}{+ 9} = - 5 \textcolor{m a \ge n t a}{+ 9} \text{ add the correct constant to both sides}$

${\left(x - 3\right)}^{2} = 4 \text{ now the LHS is a perfect square}$

$x - 3 = \pm \sqrt{4} \text{ square root both sides}$

$x = 2 + 3 , \text{ or } x = - 2 + 3$

$x = 5 , \text{ or } x = 1$

Jul 15, 2016

Another example of method. Have a look at:

https://socratic.org/s/awchNPwZ